ELEMENTARY PROBLEMS AND SOLUTIONS

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1 EEMENTARY PROBEMS AND SOUTIONS EDITED BY RUSS EUER AND JAWAD SADEK Please submit all ew problem proposals ad their solutios to the Problems Editor, DR RUSS EUER, Departmet of Mathematis ad Statistis, Northwest Missouri State Uiversity, 800 Uiversity Drive, Maryville, MO 64468, or by at All solutios to others proposals must be submitted to the Solutios Editor, DR JAWAD SADEK, Departmet of Mathematis ad Statistis, Northwest Missouri State Uiversity, 800 Uiversity Drive, Maryville, MO If you wish to have reeipt of your submissio akowledged, please ilude a self-addressed, stamped evelope Eah problem ad solutio should be typed o separate sheets Solutios to problems i this issue must be reeived by November 15, 2016 If a problem is ot origial, the proposer should iform the Problem Editor of the history of the problem A problem should ot be submitted elsewhere while it is uder osideratio for publiatio i this Joural Solvers are asked to ilude referees rather tha quotig well-kow results The otet of the problem setios of The Fiboai Quarterly are all available o the web free of harge at wwwfqmatha/ BASIC FORMUAS The Fiboai umbers F ad the uas umbers satisfy F +2 = F +1 +F, F 0 = 0, F 1 = 1; +2 = +1 +, 0 = 2, 1 = 1 Also, α = 1+ 5)/2, β = 1 5)/2, F = α β )/ 5, ad = α +β PROBEMS PROPOSED IN THIS ISSUE B-1186 Proposed by Hideyuki Ohtsuka, Saitama, Japa Prove that =0 1) = VOUME 54, NUMBER 2

2 EEMENTARY PROBEMS AND SOUTIONS B-1187 Proposed by José uis Díaz-Barrero, Bareloa Teh, Bareloa, Spai et 1 be a positive iteger Fid all real solutios of the followig system of equatios: x 3 + x+y = F 1+ )+F x 2 F y 3 +F 2 y +z = F 1+F 2 )+F 2 y2 z z +x = F 1+ 2 )+F 2z 2 B-1188 Proposed by Key B Daveport, Dallas, PA Fid a losed form expressio for 5 F k F 3 k B-1189 Proposed by Ágel Plaza, Uiversidad de as Palmas de Gra Caaria, Spai Fid a losed form for 2 ) 2 2 2k k B-1190 Proposed by José uis Díaz-Barrero, Bareloa Teh, Bareloa, Spai et 1 be a positive iteger Compute ) F +2 F +F+1 F +2 F F +1 F 1 +F+1 1 +F F +3 F +1 F F +F +1 F +2 F F +1 +F +2 F F 1 +F+1 1 +F 1 +2 F +2 +F F +1 F 1 ) +F+1 1 +F 1 +2 ) SOUTIONS Series of Reiproals B-1166 Proposed by Hideyuki Ohtsuka, Saitama, Japa Prove that F 2 k 1 k=1 2 k 2 = Solutio by Jaroslav Seibert, Uiversity of Pardubie, Czeh Republi MAY

3 THE FIBONACCI QUARTERY First we ote that 2 2 1) = 5F 2 [1, p 97], ad therefore 2 k 2 1)2k 1 = 2 k 2 = 5F 2 2 k 1 for ay positive iteger k > 1 Thus, F S = 2 k 1 k=1 2 k 2 = F F 2 k 1 k=2 2 k 2 F = 1+ 2 k 1 k=2 2 k 2 = 1+ F 2 k 1 5F 2 k=2 2 k 1 = = 4 5 F k=2 2 k F k=1 2 k 1 = F 2 k However, it is kow [1, p 437] that 1 F 2 k S = α) = = 4 α = 3+β The we have ) = Referees [1] T Koshy, Fiboai ad uas Numbers with Appliatios, Joh Wiley ad Sos, I, 2001 Also solved by Bria Bradie, Charles K Cook, Steve Edwards, Dmitry Fleishma, Russell Jay Hedel, Zbigiew Jakubzyk, Harris Kwog, ONU Problem Solvig Group, Ágel Plaza ad Fraiso Perdomo joitly), Da Weier, Nazmieh Yilmaz, ad the proposer Polygos with Fiboai Number Coordiates B-1167 Proposed by Atara Shriki ad Opher iba, Oraim College of Eduatio, Israel Fid the area of the polygo with verties F 1,F 2 ),F 3,F 4 ),F 5,F 6 ),,F 2 1,F 2 ) for 3 Solutio by Harris Kwog, SUNY Fredoia, Fredoia, NY The polygo is ovex, its iterior a be partitioed ito 2 triagles, with verties F 1,F 2 ), F 2k 1,F 2k ) ad F 2k+1,F 2k+2 ), where 2 k 1 The area k of the triagle with these three verties a be omputed by omparig the areas of the trapezoids uder its three sides ad above the x-axis) Sie F 1,F 2 ) = 1,1), we fid 180 VOUME 54, NUMBER 2

4 EEMENTARY PROBEMS AND SOUTIONS 2 k = F 2k 1 1)F 2k +1)+F 2k+1 F 2k 1 )F 2k +F 2k+2 ) F 2k+1 1)F 2k+2 +1) Therefore, = F 2k F 2k+1 F 2k 1 F 2k+2 +F 2k 1 +F 2k+2 F 2k F 2k+1 = 4k+1 1 4k F 2k = F 2k 1 1 k = 1 2 k=2 1 F 2k 1 ) 2) = F 2 2 1) 2); k=2 ad the area of the polygo is 1 2 F ) Also solved by Jeremiah Bartz, Bria D Beasley, Bria Bradie, Charles K Cook, Ravi Kumar Davala, Steve Edwards, David Farsworth, Russell Jay Hedel, Ágel Plaza ad Fraiso Perdomo joitly), Jaroslav Seibert, Jaso smith, David Stoe ad Joh Hawkis joitly), ad the proposer The Miimum Value of a Sum B-1168 Proposed by José uis Díaz-Barrero, Bareloa Teh, Bareloa, Spai et be a oegative iteger Fid the miimum value of F 2F +3 1+F ) + F 2F +3F + ) F 1+ ) Solutio by Bria Bradie, Departmet of Mathematis, Christopher Newport Uiversity, VA We make use of the followig parametrized Nesbitt s iequality [1]: et x,y,z,tx+ky+lz,ty+kz+lx,tz+kx+ly be positive real umbers ad let k l < t k+l 2 The, x tx+ky +lz + y ty+kz +lx + z tz +kx+ly 3 t+k +l, with equality if ad oly if t = k = l or x = y = z Now, let x = F, y = F, z =, t = 2, ad k = l = 3 Beause 6 = k l < 2 = t 3 = 3+3 2, it follows that F 2F +3 1+F ) + F 2F +3F + ) F 1+ ) = 3 8, for ay positive iteger, with equality if ad oly if F = = F This set of oditios holds if ad oly if = 1 Thus, the miimum value of F 2F +3 1+F ) + F 2F +3F + ) F 1+ ) MAY

5 THE FIBONACCI QUARTERY is 3 8, ad ours whe = 1 Referees [1] S Wu ad O Furdui, A ote o a ojetured Nesbitt type iequality, Taiwaese Joural of Mathematis, ), Also solved by Dmitry Fleishma, Ágel Plaza ad Fraiso Perdomo joitly), Hideyuki Ohtsuka, David Stoe ad Joh Hawkis joitly), Niuşor Zlota, ad the proposer A p-radial Iequality B-1169 Proposed by Ágel Plaza, Uiversidad de as Palmas de Gra Caaria, Spai et p be a positive iteger Prove that for ay iteger > 1 F +2 F p+1 p +1 Fp+1 F p p +F p +1 2 p+1)f 1 2 Solutio by Harris Kwog, SUNY Fredoia, Fredoia, NY We shall prove a more geeral result For ay positive real umbers x y, ad for ay positive iteger p, we shall prove that ) x+y p xp+1 y p+1 p p+1)x y) xp +y p 1) 2 The desired result will follow if we set x = F +1 ad y = F It is straightforward to verify that p p+1)x p +y p ) 2 x p i y i i=0 p 1 = p 1)x p +y p ) 2 x p i y i 2 = p 1)/2 i=1 x p/2 y p/2 ) i=1 x p i y p i )x i y i ) if p is odd, p 1)/2 i=1 x p i y p i )x i y i ) if p is eve This proves the seod iequality i 1) By settig t = x/y, it is lear that i order to establish the first iequality i 1), it suffies to prove that, for ay iteger p 1, f p t) = 2 p t p+1 1) p+1)t 1)t+1) p VOUME 54, NUMBER 2

6 EEMENTARY PROBEMS AND SOUTIONS wheever t 1 We shall proeed by idutio o p Our laim is valid whe p = 1 beause f 1 t) = 0 Assume that f p t) 0 wheever t 1 for some iteger p 1 We fid f p+1t) = 2 p+1 p+2)t p+1 p+2)t+1) p+1 p+2)p+1)t 1)t+1) p = p+2)[2 p t p+1 1) p+1)t 1)t+1) p +2 p t p+1 +1) t+1) p+1 ] = p+2)[f p t)+2 p t p+1 +1) t+1) p+1 ] For t 1, we gather from the idutio hypothesis that f p t) 0, ad from the power mea iequality we dedue that p+1 t p+1 +1 t+1 2 2, whih implies that 2 p t p+1 +1) t+1) p+1 Hee, f p+1 t) 0, ad f p+1t) is o-dereasig whe t 1 Together with f p+1 1) = 0, we olude that f p+1 t) 0 wheever t 1 This ompletes the idutio ad the proof of the first iequality i 1) Also solved by Dmitry Fleishma, Wei-Kai ei, Hideyuki Ohtsuka, Niuşor Zlota, ad the proposer imits of Radials Equality B-1170 Proposed by D M Bătieţu Giurgiu, Matei Basarab Natioal College, Buharest, Romaia ad Neulai Staiu, George Emil Palade Shool, Buzău, Romaia et a > 0, b > 0, ad 0 Fid i) ii) lim lim +1) )!aF+1 +1) )!a +1 2!bF, 2!b Solutio by Hideyuki Ohtsuka Saitama, Japa), Ágel Plaza ad Fraiso Perdomo Uiversidad de as Palmas de Gra Caaria, Spai), Jaroslav Seibert Uiversity of Pardubie, Czeh Republi), idepedetly emma 1 [1] If the positive sequees {p } ad {q } are suh that p +1 q lim = p > 0 ad = q > 0, p p the, lim +1 p +1 q ) = p e log e q MAY

7 THE FIBONACCI QUARTERY i) et We have p = 2!aF p +1 1 lim = lim p = lim = 1 α lim ad q = 2!bF +1) 2+1) +1)!aF+1!aF 2 ) ) F 1+ 1 F +1 ) 2 ) 1+ 1 ad Usig the lemma, we have lim +1) )!aF+1 = e2 α 2!bF q p = a b = lim ) +1) 2+1) +1)!aF+1 2!bF = lim +1 p +1 q ) = e eb log α a ii) Similarly, we have lim +1) 2 2 = e eb log +1 +1)!a!b +1 α a Referees [1] Gh Toader, alesu sequees, Publikaije-Elektrotehikog Fakulteta Uiverzitet U Beogradu Serija Matematika, ), Also solved by Dmitry Fleishma ad the proposer We would like to belatedly akowledge the solutios to Problems B-1161 through B-1165 by Dmitry Fleishma VOUME 54, NUMBER 2